The variable is passed through ajax and it is a database name. When the link is clicked it's supposed to retrieve a data from the the database. the link and the variable is on the same page. Here is the code for the link:
$x = strval($_GET['x']);
echo '<a href="#" onclick="showInformation('.$x.')">'.$seatid.'</a>';
The $x variable contains the table name of the database. And here is the code for ajax:
function showInformation(str)
{
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtInfo").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getinfo.php?x="+str,true);
xmlhttp.send();
}
Here is the getinfo.php:
<?php
session_start();
$_SESSION['login']="1";
$x = strval($_GET['x']);
$con = mysql_connect('localhost','root','Newpass123#','seatmapping');
if (!$con){
die('Could not connect: ' . mysql_error());
}
mysql_select_db('seatmapping');
$sql="SELECT name, seatid FROM $x WHERE seatid = 1";
$result = mysql_query($sql) or die("Query Error " . mysql_error());
...
...
?>
I can't get it to work when i click the link it does not display the data from the table. Please help me. Any kind of help is appreciated. Thanks in advance..
Can you please try this:
Replace the line
echo '<a href="#" onclick="showInformation('.$x.')">'.$seatid.'</a>';
With the below:
echo '<a href="#" onclick="showInformation(\''.$x.'\')">'.$seatid.'</a>';
This will solve your problem.
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